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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
`log("cosec x"-cot x)` |
| Answer» Correct Answer - `"cosec x"` | |
| 52. |
Differentiate with respect to x:√(sin x) |
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Answer» Chain rule: If y = f(t) and t = g(x) then dy/dx = dy/dt x dt/dx d/dx √(sin x) = d/dx (sin x)1/2 = 1/2 * (sin x)-1/2 * d/dx (sin x) = 1/2 * (sin x)-1/2 * cos x = cos x/2√(sin x) |
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| 53. |
Find `(dy)/(dx)`, if `x=a(theta+sintheta), y=1(1-costheta)`. |
| Answer» Correct Answer - `cot.(theta)/(2)` | |
| 54. |
Find `(dy)/(dx),`when`x=a(theta+sintheta)a n dy=a(1-costheta)` |
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Answer» `x = a(theta+sintheta)` `:. dx/(d theta) = a(1+cos theta)` `y = a(1-cos theta)` `:. dy/( d theta) = asin theta` `:. dy/dx = (dy/( d theta))/(dx/( d theta)) = ( asin theta)/(a(1+cos theta))` `=>dy/dx = sin theta/(1+cos theta) = (2sin(theta/2)cos (theta/2))/(2cos^2(theta/2)) = tan(theta/2)` `=>:. dy/dx = tan(theta/2)`. |
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| 55. |
Find `(dy)/(dx),`when`x=a e^(theta)(sintheta-costheta),y=a e^(theta)(sintheta+costheta)` |
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Answer» `x = ae^theta(sintheta-costheta)` `:. dx/(d theta) = ae^(theta)(cos theta+sin theta)+ae^theta(sintheta-cos theta) = 2ae^thetasintheta` `y = ae^theta(sintheta-costheta)` `:. dy/( d theta) =ae^(theta)(cos theta-sin theta)+ae^theta(sintheta+cos theta) = 2ae^thetacostheta` `:. dy/dx = (dy/( d theta))/(dx/( d theta)) = (2ae^thetacostheta)/(2ae^thetasintheta)` `=> dy/dx = cot theta`. |
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| 56. |
Find `(dy)/(dx)`, when: `y=sin2x sin 3x sin 4x` |
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Answer» Correct Answer - `(sin2x sin 3x sin 4x)(2cot 2x+3 cot 3x+4 cot 4x)` `logy=logsin2x+log sin 3x+log sin 4x` `rArr(1)/(y).(dy)/(dx)={(2cos2x)/(sin 2x)+(3cos3x)/(sin3x)+(4 cos 4x)/(sin 4x)}.` |
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| 57. |
Find `(dy)/(dx)ifx=a(theta-sintheta)a n dy=a(1-costheta)dot` |
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Answer» `"We have "x=a(theta-sin theta)and y=a (1- cos theta).` `therefore" "(dx)/("d"theta)=a(1-cos theta) and (dy)/("d"theta)=a sin theta` `"or "(dy)/(dx)=(dy//"d" theta)/(dx//"d" theta)` `=(a sin theta)/(a(1-cos theta))` `=(2sin (theta//2)cos (theta//2))/(2 sin^(2) (theta//2))` `=cot""(theta)/(2)` |
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| 58. |
If `y=y(x)`and itfollows the relation `4x e^(x y)=y+5sin^2x ,`then `y^(prime)(0)`is equalto______ |
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Answer» We have `4xe^(xy)=y+5 sin^(2)x` Putting, x = 0 in above equation we get y=0 `therefore" (0,0) lies on the curve"` Now on differentaiting equation (1) w.r.t.x, we get `4ee^(xy)+4xe^(xy)(x(dy)/(dx)+y)=(dy)/(dx)+10 sin x cos x` Putting x= 0 and y=0, we get `4+0=((dy)/(dx))_("(0,0)")+0` `therefore" "((dy)/(dx))_("(0,0)")=4` |
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| 59. |
If`y=sqrt(x^2+a^2)` ,prove that `y(dy)/(dx)-x=0` |
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Answer» `y=sqrtx` `=dy/dx` `=1/(2sqrtx)` `dy/dx=(1xx2x)/(2(sqrtx^2+a^2)` `dy/dx=x/(sqrt(x^2+a^2))` `y=sqrt(a^2+x^2)` `(dy/dx)y-x=0` `=x/sqrt(x^2+a^2)sqrt(a^2+x^2)-x` `=0=RHS` |
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| 60. |
`"If "log(x^(2)+y^(2))=2tan^(-1)""((y)/(x))," show that "(dy)/(dx)=(x+y)/(x-y)` |
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Answer» Differentiating both sides w.r.t. x, we get `(d)/(dx){log(x^(2)+y^(2))}=2(d)/(dx){tan^(-1)((y)/(x))}` `"or "(1)/(x^(2)+y^(2))xx(d)/(dx)(x^(2)+y^(2))=2(1)/(1+(y//x)^(2))xx(d)/(dx)((y)/(x))` `"or "(1)/(x^(2)+y^(2)){(d)/(dx)(x^(2))+(d)/(dx)(y^(2))}=2xx(x^(2))/(x^(2)+y^(2)){(x(dy)/(dx)-yxx1)/(x^(2))}` `"or "(1)/(x^(2)+y^(2)){2x+22y(dy)/(dx)}=(2)/(x^(2)+y^(2)){x(dy)/(dx)-y}` `"or "{x+y(dy)/(dx)}=2{x(dy)/(dx)-y}` `"or "x+y(dy)/(dx)=x(dy)/(dx)-y` `"or "(dy)/(dx)(y-x)=-(x+y)` `"or "(dy)/(dx)=(x+y)/(x-y)` |
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| 61. |
If`y=sin[2tan^(-1){sqrt((1-x)/(1+x))}],`find`(dy)/(dx)` |
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Answer» `y=sin[2tan^-1 (sqrt(1-x)/(1+x))]` `x=Costheta` `y=sin[2tan^-1 (sqrt(1-Costheta)/(1+Costheta))]` `y=sin[2tan^-1 (sqrt(2-Sin^2theta/2)/(1+Cos^2theta/2))]` `y= sin[2tan^-1tan(theta/2)]` `y=Sintheta` `x=Costheta` `(dy)/(d theta)=costheta=x` `(dx)/(d theta)=-sintheta=-y` `(dy)/(dx)=(dy)/(d theta).(d theta)/(dx)=(x)(-1)/(y)` `(dy)/(dx)=-x/y` |
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| 62. |
If `y=xlog"{"x/((a+b x))"]"`, then show that`x^3(d^2y)/(dx^2)=(x(dy)/(dx)-y)^2dot` |
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Answer» Given `y//x=[log x- log (a+bx)].` Therefore, `(1)/(x)(dy)/(dx)-(1)/(x^(2))y=(1)/(x)-(b)/(a+bx)" [Diff. both sides w.r.t. x]"` `" or "x(dy)/(dx)-y=(ax)/(a+bx)" (1)"` Differentiating again w.r.t. x, we get `(x(d^(2)y)/(dx^(2))+(dy)/(dx))-(dy)/(dx)=(a^(2))/(a+bx)^(2)` `therefore" "x^(3)(d^(2)y)/(dx^(2))=(a^(2)x^(2))/((a+bx)^(2))=(x(dy)/(dx)-y)^(2)` |
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| 63. |
If`y=sqrt((1-x)/(1+x)),p rov et h a t(1-x^2)(dy)/(dx)+y=0` |
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Answer» We have `y=sqrt((1-x)/(1+x))` Differentiating w.r.t.x, we get `(dy)/(dx)=(1)/(2)((1-x)/(1+x))^((1//2)-1)(d)/(dx)((1-x)/(1+x))` `=(1)/(2)sqrt((1+x)/(1-x))((1+x)(d)/(dx)(1-x)-(1-x)(d)/(dx)(1+x))/(1+x)^(2)=-sqrt((1+x)/(1-x))(1)/((1+x)^(2))` `"or " (1-x^(2))(dy)/(dx)=-sqrt((1+x)/(1-x))(1)/((1+x)^(2))(1-x^(2))` `"or " (1-x^(2))(dy)/(dx)=-sqrt((1-x)/(1+x))` `"or " (1-x^(2))(dy)/(dx)=-y` `"or " (1-x^(2))(dy)/(dx)+y=0` |
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| 64. |
Find the sum of the series `1+2x+3x^(2)+(n-1)x^(n-2))` using differentiation. |
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Answer» We know that `1+x+x^(2)+…+x^(n-1)=(1-x^(n))/(1-x).` Differentiating both sides w.r.t.x, we get `0+1+2x+3x^(2)+…+(n-1)x^(n-2)` `=((1-x)(d)/(dx)(1-x^(n))-(1-x^(n))(d)/(dx)(1-x))/((1-x)^(2))` `"or "1+2x+3x^(2)+...+(n-1)^(x-2)=(-(1-x)nx^(n-1)+(1-x^(n)))/((1-x)^(2))` `"or "1+2x+3x^(2)+...+(n-1)^(x-2)=(-nx^(n-1)+(n-1)x^(n)+1)/((1-x)^(2))` |
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| 65. |
\(y = (x - 1) (5x^3 + \sqrt x)^\frac{11}2 \left(\frac{tanx - 1}{sin x}\right) e^{{5(x)}^\frac18} .\, x^{x^{x....}}\) |
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Answer» \(y = (x - 1) (5x^3 + \sqrt x)^\frac{11}2 \left(\frac{tanx - 1}{sin x}\right) e^{{5(x)}^\frac18} .\, x^{x^{x....}}\) \(log \,y = log (x -1) + \frac{11}2 log (5x^3 + \sqrt x) + log(tan x - 1) - log \, sin x + 5(x) \frac 18 + x^{x^x} \) On Differentiating both sides logx w.r.t x, we get \(\frac1y \frac {dy}{dx} = \frac 1 {x- 1} +\frac{11}2\frac{15x^2 + \frac 1{2\sqrt x}}{5x^3+ \sqrt x} + \frac{sec^2 x}{tan x - 1} - \frac{cos x}{sin x} + \frac 58 x^{-\frac78} + \frac{x^{x^{x...}}}{x} + log x. \frac{x^{x^{x...}}}{x}\left(\frac 1{x^{x^{x...} }} - log x\right)\) \(\therefore \frac{dy}{dx} = y\left(\frac 1{x - 1} + \frac {11}2 \frac{15x^2 + \frac1{2\sqrt x}}{5x^3 + \sqrt x} + \frac{sec^2x}{tan x -1} - cot x + \frac 58 \, \frac1{x^\frac 78} + \frac{x^{x^{x...}}}{x}\left(\frac {log x}{x^{x^{x..}}} - (log x)^2 + 1\right) \right)\) y at x = 1 = 0 \(\therefore\frac{dy}{dx}|_{x=1} = 0\) |
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| 66. |
A curve is represented parametrically by the equations `x=f(t)=a^(In(b^t))and y=g(t)=b^(-In(a^(t)))a,bgt0 and a ne 1, b ne 1" Where "t in R.` The value of `(d^(2)y)/(dx^(2))` at the point where f(t)=g(t) is |
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Answer» `x=f(t)=a^(In(b))=a^(t" In "b)" (1)"` `y=g(t)=b^(-In(a^(t)))=(b^(In a))^(-t)=(a^(In b))^(-t)=a^(-t" In b")` `therefore" "y=g(t)=a^(In(b^(-t)))=f(-t)" (2)"` From equations (1), and (2), `xy=1` `f(t)=g(t)rArrf(t)=f(-t)rArrt=0" "[because f(t)" is one-one function"]` `"At "t=0, x=y=1` `because" "xy=1, (dy)/(dx)=(-1)/(x^(2))and (d^(2)y)/(dx^(2))=(2)/(x^(3))" At "x=1, (d^(2)y)/(dx^(2))=2` |
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| 67. |
A curve parametrically given by `x=t+t^(3)" and "y=t^(2)," where "t in R." For what vlaue(s) of t is "(dy)/(dx)=(1)/(2)`?A. `1//3`B. 2C. 3D. 1 |
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Answer» `(dy)/(dx)=(dy//dt)/(dx//dt)=(2t)/(1+3t^(2))=(1)/(2)`(given) `rArr" "3t^(2)-4t+1=0` `rArr" "(3t-1)(t-1)=0` `rArr" "t=(t)/(3),t=t` |
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| 68. |
If `y=e^(-x) cos x` and `y_n+k_ny=0` where `yn=(d^ny)/(dx^n)` and `k_n` are constant `n in N` thenA. `k_(4)=4`B. `k_(8)=-16`C. `k_(12)=20`D. `k_(16)=-24` |
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Answer» Correct Answer - B `y=e^(-x)cos x` `y_(1)=-e^(-x)cos x - e^(-x)sin x=-sqrt(2) e^(-x)cos (x-(pi)/(4))` `y_(2)=(-sqrt(2))e^(-x)cos (x-(pi)/(2))` `y_(3)=(-sqrt(2))^(3)e^(-x)cos (x-(3pi)/(4))` `y_(4)=(-sqrt(2))^(4)e^(-x)cos (x-pi)=-4 e^(-x)cos x` `"or "y_(4)+4y=0or k_(4)=4` Differentiating it again four times, we get `y_(8)+4y_(4)=0` `"or "y_(8)-16y=0` `"or "k_(8)=-16` `"Further "y_(12)+4y_(8)=0` `"or "y_(12)+64y=0` `"or "k_(12)=64` `"Similarly, "k_(16)=-256` |
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| 69. |
If`y=sqrt(logx+sqrt(cosx+sqrt(cosx+...tooo))),`provethat `(dy)/(dx)=(sinx)/(1-2y)` |
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Answer» `y = sqrt(cosx+sqrt(cosx+sqrt(cosx+...oo)))` `=>y = sqrt(cosx+y)` `=>y^2 = cosx+y` Differntiating both sides w.r.t. `x`, `=>2y dy/dx = -sinx+dy/dx` `=>dy/dx(2y-1) = -sinx` `=>dy/dx = sinx/(1-2y)` |
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| 70. |
If `y=(logx)^(cosx)+(x^(2)+1)/(x^(2)-1)`, find `(dy)/(dx).` |
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Answer» Let `(logx)^(cosx)=u and(x^(2)+1)/(x^(2)-1)=v.` Then, `u=(logx)^(cosx)rArr log u = cosx. Log (logx)." …(i)"` On differentiating both sides of (i) w.r.t. x, we get `(1)/(u).(du)/(dx)=cosx.(d)/(dx){log(logx)}+log(logx).(d)/(dx)(logx)` `=cosx(1)/(logx).(1)/(x)-(sinx).log(logx).` `therefore(du)/(dx)=u.{(cosx)/(xlogx)-(sinx).log(logx)}` `rArr(du)/(dx)=(logx)^(cosx).{(cosx)/(x logx)-(sinx).log(logx)}.` And, `v=((x^(2)+1))/((x^(2)-1))rArr(dv)/(dx)=((x^(2)-1).(d)/(dx)(x^(2)+1)-(x^(2)+1).(d)/(dx)(x^(2)-1))/((x^(2)-1)^(2))` `rArr(dv)/(dx)=((x^(2)-1).2x-(x^(2)+1).2x)/((x^(2)-1)^(2))=(-4x)/((x^(2)-1)^(2)).` `therefore y=u+v` `rArr(dy)/(dx)=(du)/(dx)+(dv)/(dx)` `=(logx)^(cosx).{(cosx)/(xlogx)-(sinx).(logx)}-(4x)/((x^(2)-1)^(2)).` |
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| 71. |
Find `(dy)/(dx)`, when: `y=(cosx)^(logx)` |
| Answer» Correct Answer - `(cosx)^(logx).{(log(cosx))/(x)-(tanx)(logx)}` | |
| 72. |
Find `(dy)/(dx)`for the function:`y=a^(sin^(-1)x)^2` |
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Answer» Correct Answer - `(2log a sin^(-1)x)/(sqrt(1-x^(2)))a(sin^(-1)x)^(2)` `"Let "y=a^((sin^(-1)x)^(2))`. Using chain rule, we get `(dy)/(dx)=(d)/(dx){a^((sin^(-1)x)^(2))}` `=a^((sin^(-1)x)^(2))loga(d)/(x){(sin^(-1)x)^(2)}` `=a^((sin^(-1)x)^(2))(log a) 2 (sin ^(-1)x)^(1)(d)/(dx)(sin^(-1)x)` `=a^((sin^(-1)x)^(2))(log a)2 sin^(-1)x(1)/(sqrt(1-x^(2)))` `=(2 log a sin^(-1)x)/(sqrt(1-x^(2)))a^((sin^(-1)x)^(2))` |
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| 73. |
`y=sin^(-1)[sqrt(x-a x)-sqrt(a-a x)]` |
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Answer» Correct Answer - `(1)/(2sqrt(x(1-x)))` `y=sin^(-1)[sqrt(x-ax)-sqrt(a-ax)]` `=sin^(-1)[sqrt(x)sqrt(1-a)-sqrt(a)sqrt(1-x)]` `=sin^(-1)[sqrt(x)sqrt(1-(sqrt(a))^(2))-sqrt(a)sqrt(1-(sqrt(x))^(2))]` `=sin^(-1)sqrt(x)-sin^(-1)sqrt(a)` `therefore" "(dy)/(dx)=(1)/(sqrt(1-x))xx(1)/(2sqrt(x))` |
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| 74. |
Find `(dy)/(dx)`if `y=log{e^x((x-2)/(x+2))^(3/4)}` |
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Answer» Correct Answer - `(x^(2)-1)/(x^(2)-4)` `"Let "y=[log{e^(x)((x-2)/(x+2))^(3//4)}]=log e^(x)+log ((x-2)/(x+2))^(3//4)` `=x+(3)/(4)[log (x-2)-log(x+2)]` `therefore" "(dy)/(dx)=1+(3)/(4)[(1)/(x-2)-(1)/(x+2)]=1+(3)/((x^(2)-4))=(x^(2)-1)/(x^(2)-4)` |
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| 75. |
Differentiate the following functions:(i) 3x-5 (ii) 1/5x (iii) 6(x2)3/2 |
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Answer» We know, d/dx (xm) = m(xm-1) (i) d/dx (3x-5) = 3(-5)x-6 = -15 x-6 (ii) d/dx (1/5x) = 1/5 * (d/dx (1/x)) = 1/5 (-1/x2) = -1/5x2 (iii) d/dx (6 (x2)1/3) = 6 d/dx (x2/3) = 6(2/3 * x-1/3) = 4x-1/3 |
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| 76. |
Differentiate `(e^(x)cos^(3)x sin^(2)x)` w.r.t. x. |
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Answer» Let `y=e^(x)cos^(3)xsin^(2)x." …(i)"` Taking logarithm on both sides of (i), we get `logy=x+3 log cos x+2 log sin x." …(ii)"` On differentiating both sides of (ii) w.r.t. x, we get `(1)/(y).(dy)/(dx)=1+(3)/(cosx).(-sinx)+(2)/(sinx).cosx` `rArr(dy)/(dx)=y.{1-3 tanx+2 cot x}` `=(e^(x)cos^(3)xsin^(2)x)(1-3tanx+2cotx).` |
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| 77. |
Differentiate `sqrt((x-1)(x-2)(x-3)(x-4))` w.r.t. x. |
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Answer» Let `y=sqrt((x-1)(x-2)(x-3)(x-4))." …(i)"` Taking logarithm on both sides of (i), we get `log y = (1)/(2){log(x-1)+logx(x-2)+log(x-3)+log(x-4)}……..(ii)` On differentiating both sides of (ii) w.r.t. x we get `(1)/(y).(dy)/(dx)=(1)/(2).{(1)/((x-1))+(1)/((x-2))+(1)/((x-3))+(1)/((x-4))}` `(dy)/(dx)=((y)/(2)).{(1)/((x-1))+(1)/((x-2))+(1)/((x-3))+(1)/((x-4))}` `=(1)/(2).sqrt((x-1)(x-2)(x-3)(x-4)).{(1)/((x-1))+(1)/((x-2))+(1)/((x-3))+(1)/((x-4))}` |
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| 78. |
Differentiate `(x+1)^(2)(x+2)^(3)(x+3)^(4)` w.r.t. x. |
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Answer» Let `y=(x+1)^(2)(x+2)^(3)(x+3)^(4)." …(i)"` Taking logarithm on both sides of (i), we get `log y = 2log (x+1)+3log (x+2)+4 log (x+3)." …(ii)"` Differentiating both sides of (ii) w.r.t. x, we get `(1)/(y).(dy)/(dx)=(2)/((x+1))+(3)/((x+2))+(4)/((x+3))` `rArr(dy)/(dx)=y.[(2)/((x+1))+(3)/((x+2))+(4)/((x+3))]` `=(x+1)^(2)(x+2)^(3)(x+3)^(4).[(2)/((x+1))+(3)/((x+2))+(4)/((x+3))].` |
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| 79. |
If `y=(sqrtx(x+4)^(3//2))/((4x-3)^(4//3)),` find `(dy)/(dx).` |
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Answer» Given: `y=(sqrtx(x+4)^(3//2))/((4x-3)^(4//3))." ...(i)"` Taking logarithm on both sides of (i), we get `log y = (1)/(2)log x+(3)/(2)log (x+4)-(4)/(3)log(4x-4).` On differentiating both sides w.r.t. x, we get `(1)/(y).(dy)/(dx)=(1)/(2).(1)/(x)+(3)/(2).(1)/((x+4))-(4)/(3).(4)/((4x-3))` `rArr(dy)/(dx)=y[(1)/(2x)+(3)/(2(x+4))-(16)/(3(4x-3))]` `=(sqrtx(x+4)^(3//2))/((4x-3)^(4//3)).[(1)/(2x)+(3)/(2(x+4))-(16)/(3(4x-3))].` |
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| 80. |
If `y=sqrt(((x-3)(x^(2)+4))/((3x^(2)+4x+5)))`, find `(dy)/(dx)`. |
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Answer» Given : `y=sqrt(((x-3)(x^(2)+4))/((3x^(2)+4x+5))).` Taking logarithm on both sides of (i), we get `logy=(1)/(2){log(x-3)+log(x-3)+log(x^(2)+4)-log(3x^(2)+4x+5)}.` Differentiating both sides w.r.t. x, we get `(1)/(y).(dy)/(dx)=(1)/(2).{(1)/((x-3))+(2x)/((x^(2)+4))-((6x+4))/((3x^(2)+4x+5))}` `rArr (dy)/(dx)=((1)/(2)y).{(1)/((x-3))+(2x)/((x^(2)+4))-((6x+4))/((3x^(2)+4x+5))}` `=(1)/(2).sqrt(((x-3)(x^(2)+4))/((3x^(3)+4x+4))).{(1)/((x-3))+(2x)/((x^(2)+4))-((6x+4))/((3x^(2)+4x+5))}` |
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| 81. |
If `tan^(-1)((x^(2)-y^(2))/(x^(2)+y^(2)))=a`, show that `(dy)/(dx)=(x(1-tana))/(y(1+tana)).` |
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Answer» `tan^(-1)((x^(2)-y^(2))/(x^(2)+y^(2)))=arArr((x^(2)-y^(2))/(x^(2)+y^(2)))=tana` `therefore (x^(2)-y^(2))=(x^(2)+y^(2)) tan a." …(i)"` On differentiating both sides of (i) w.r.t. x, we get `2x-2y.(dy)/(dx)=2x tan a +2y.(dy)/(dx).tana` `rArr y(1+tana)(dy)/(dx)=x(1-tana)` `rArr (dy)/(dx)=(x(1-tana))/(y(1+tana)).` |
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| 82. |
If`e^x+e^y=e^(x+y),` prove that `(dy)/(dx)+e^(y-x)=0` |
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Answer» Given : `e^(x)+e^(y)=e^(x+y)." …(i)"` On dividing throughout by `e^(x+y)`, we get `e^(-y)+e^(-x)=1." …(ii)"` `e^(-y).((-dy)/(dx))+e^(-x)(-1)=0` `rArr(dy)/(dx)=(-e^(-x))/(e^(-y))=-e^((y-x)).` |
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| 83. |
If `sin(x y)+y/x=x^2-y^2`, find `(dy)/(dx)`. |
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Answer» Given: `sin(xy)+(x)/(y)=x^(2)-y.` Differentiating both sides w.r.t. x, we get `cos(xy).(d)/(dx)(xy)+x.(-(1)/(y^(2)))(dy)/(dx)+(1)/(y).1=2x-(dy)/(dx)` `rArr cos(xy).[x.(dy)/(dx)+y.1]-(x)/(y^(2)).(dy)/(dx)+(1)/(y)=2x-(dy)/(dx)` `rArr[x cos(xy)-(x)/(y^(2)+1)].(dy)/(dx)=2x-(1)/(y)-y cos(xy)` `rArr{xy^(2)cos(xy)-x+y^(2)}.(dy)/(dx)=2xy^(2)-y-y^(3)cos(xy).` Hence, `(dy)/(dx)={(2xy^(2)-y-y^(3)cos(xy))/(xy^(2)cos(xy)-x+y^(2))}.` |
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| 84. |
If `cos(x+y)=y sin x,` find `(dy)/(dx).` |
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Answer» Given: `cos(x+y)=y sin x.` On differentiating both sides of (i) w.r.t. x, we get `-sin(x+y).(d)/(dx)(x+y)=y cos+sinx.(dy)/(dx)` `rArr-sin(x+y)(1+(dy)/(dx))=y cos x+sin x.(dy)/(dx)` `rArr{sin(x+y)+sinx}.(dy)/(dx)=-{sin(x+y)+ycosx}` `rArr(dy)/(dx)=(-{sin(x+y)+ycosx})/({sin(x+y)+sinx}).` |
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| 85. |
If `y=sqrtx+1/sqrtx` then prove that `2x (dy)/(dx)+y=2sqrtx` |
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Answer» `y=sqrtx+(1)/(sqrtx)rArrsqrtxy=x+1" ...(i)"` On differentiating both sides of (i) w.r.t. x, we get `sqrtx.(dy)/(dx)y.(1)/(2)x^(-1//2)=1` `rArrsqrtx(dy)/(dx)+(y)/(2sqrtx)=1` `rArr 2x(dy)/(dx)+y=2sqrtx.` |
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| 86. |
`(i)e^(2//x)" "(ii)e^(sqrtx)" "(iii) e^(-2sqrtx)` |
| Answer» `(i)-(2)/(x^(2)).e^(2//x)" "(ii)(1)/(2sqrtx)e^(sqrtx)" "(iii)(-1)/(sqrtx)e^(-2sqrtx)` | |
| 87. |
`y=sin^(-1)""(2x)/(1+x^(2)),-1lexle1` |
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Answer» Correct Answer - `(2)/(1+x^(2))` `y=sin^(-1)((2x)/(1+x^(2)))` Let `x = tan theta.` Therefore, `y=sin^(-1)((2tan theta)/(1+tan^(2)theta))` `=sin^(-1)(sin 2 theta)` `=2theta` `=2tan^(-1)x` `therefore" "(dy)/(dx=(d)/(dx)(2 tan^(-1)x)` `=2(d)/(dx)(tan^(-1)x)` `(2)/(1+x^(2))` |
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| 88. |
`sin^(-1){(1)/(sqrt(1+x^(2)))}` |
| Answer» Correct Answer - `(1)/((1+x^(2)))` | |
| 89. |
`tan^(- 1)((1+x)/(1-x))=pi/4+tan^(- 1)x , xlt1` |
| Answer» Correct Answer - `(1)/((1+x^(2)))` | |
| 90. |
`sec^(-1)((1)/(1-2x^(2)))` |
| Answer» Correct Answer - `(2)/(sqrt(1-x^(2)))` | |
| 91. |
`sec^(- 1)((x^2+1)/(x^2-1))` |
| Answer» Correct Answer - `(-2)/((1+x^(2)))` | |
| 92. |
`"cosec"^(-1)((1+x^(2))/(2x))` |
| Answer» Correct Answer - `(2)/((1+x^(2)))` | |
| 93. |
`tan^(-1)((3x-x^(3))/(1-3x^(2)))` |
| Answer» Correct Answer - `(3)/((1+x^(2)))` | |
| 94. |
`sec^(-1)((1+x^(2))/(1-x^(2)))` |
| Answer» Correct Answer - `(2)/((1+x^(2)))` | |
| 95. |
`sec^(-1)((1+tan^(2)x)/(1-tan^(2)x))` |
| Answer» Correct Answer - 2 | |
| 96. |
The differential coefficient of `a^(log10" cosec"^(-1)x)`, isA. `(a^(log10("cosec"^(-1)x)))/("cosec"^(-1)x)(1)/(xsqrt(x^(2)-1))log_(10)a`B. `-(a^(log10("cosec"^(-1)x)))/("cosec"^(-1)x).(1)/(|x|sqrt(x^(2)-1))log_(10)a`C. `(-a^(log10("cosec"^(-1)x)))/("cosec"^(-1)x).(1)/(|x|sqrt(x^(2)-1))log_(a)10`D. `(a^(log10"cosec"^(-1)x))/("cosec"^(-1)x).(1)/(xsqrt(x^(2)-1))log_(a)10` |
| Answer» Correct Answer - B | |
| 97. |
Differentiate with respect to x:1/(x2 - x + 3)3 |
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Answer» Chain rule: If y = f(t) and t = g(x) then dy/dx = dy/dt x dt/dx d/dx (x2 – x + 3)-3 = -3(x2 – x + 3)-4 * d/dx (x2 – x + 3) = -3(x2 – x + 3)-4 (2x2 – 1) |
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| 98. |
Differentiate with respect to x:sin2 (2x + 3) |
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Answer» Chain rule: If y = f(t) and t = g(x) then dy/dx = dy/dt x dt/dx d/dx sin2 (2x + 3) = 2 sin (2x + 3) * d/dx {sin(2x + 3)} = 2 sin (2x + 3) * cos(2x + 3) * d/dx (2x + 3) = 2 sin (2x + 3) * cos (2x + 3) * 2 = 4 sin (2x + 3) cos (2x + 3) = 2 sin (4x + 6) [Using identity: 2sin x cos x = sin 2x] |
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| 99. |
(d2 - 3d + 2)y = 0, where D = d/dx is _____ |
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Answer» (D2 – 3D + 2)y = 0 It's auxilarly equation is m2 – 3m + 2 = 0 ⇒ m2 – 2m – m + 2 = 0 ⇒ m(m – 2) – 1(m – 2) = 0 ⇒ (m – 2)(m – 1) = 0 ⇒ m = 1, 2 Therefore, its solution is y = C1ex + C2e2x. |
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| 100. |
Differentiate using chain rule cos2(x3) |
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Answer» Chain Rule: If y = f(t) and t = g(x) then dy/dx = dy/dt x dt/dx d/dx cos2(x3) = 2 cos(x3) * d/dx {cos(x3)} = 2 cos(x3) * ( – sin (x3)) * d/dx (x3) = 2 cos(x3) * ( – sin (x3)) * 3x2 = – 6x2 cos(x3) sin x3 |
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